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Difference between revisions of "2019 AIME II Problems/Problem 9"

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If <math>a+1 = 4</math>, then <math>b+1 = 5</math>. But this leads to no solutions, as <math>(a,b) = (3,4)</math> gives <math>2^3 5^4 > 2019</math>.
 
If <math>a+1 = 4</math>, then <math>b+1 = 5</math>. But this leads to no solutions, as <math>(a,b) = (3,4)</math> gives <math>2^3 5^4 > 2019</math>.
  
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If <math>a+1 = 5</math>, then <math>b+1 = 2</math> or <math>4</math>. The first case gives <math>n = 2^4 \cdot 5^1 \cdot p</math> where <math>p</math> is a prime other than 2 or 5. Thus we have <math>80p < 2019 \implies p = 3, 7, 11, 13, 17, 19, 23</math>. The sum of all such <math>n</math> is <math>80(3+7+11+13+17+19+23) = 5760</math>. In the second case <math>b+1 = 4</math> and <math>d(k) = 1</math>, and there is one solution <math>n = 2^4 \cdot 5^3 = 2000</math>.
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If <math>a+1 = 5</math>, then <math>b+1 = 2</math> or <math>4</math>. The first case gives <math>n = 2^4 \cdot 5^1 \cdot p</math> where <math>p</math> is a prime other than 2 or 5. Thus we have <math>80p < 2019 \implies p = 3, 7, 11, 13, 17, 19, 23</math>. The sum of all such <math>n</math> is <math>80(3+7+11+13+17+19+23) = 7440</math>. In the second case <math>b+1 = 4</math> and <math>d(k) = 1</math>, and there is one solution <math>n = 2^4 \cdot 5^3 = 2000</math>.
  
 
If <math>a+1 = 10</math>, then <math>b+1 = 2</math>, but this gives <math>2^9 \cdot 5^1 > 2019</math>. No other values for <math>a+1</math> work.
 
If <math>a+1 = 10</math>, then <math>b+1 = 2</math>, but this gives <math>2^9 \cdot 5^1 > 2019</math>. No other values for <math>a+1</math> work.

Revision as of 16:53, 7 March 2020

Problem 9

Call a positive integer $n$ $k$-pretty if $n$ has exactly $k$ positive divisors and $n$ is divisible by $k$. For example, $18$ is $6$-pretty. Let $S$ be the sum of positive integers less than $2019$ that are $20$-pretty. Find $\tfrac{S}{20}$.

Solution

Every 20-pretty integer can be written in form $n = 2^a 5^b k$, where $a \ge 2$, $b \ge 1$, $\gcd(k,10) = 1$, and $d(n) = 20$, where $d(n)$ is the number of divisors of $n$. Thus, we have $20 = (a+1)(b+1)d(k)$, using the fact that the divisor function is multiplicative. As $(a+1)(b+1)$ must be a divisor of 20, there are not many cases to check.

If $a+1 = 4$, then $b+1 = 5$. But this leads to no solutions, as $(a,b) = (3,4)$ gives $2^3 5^4 > 2019$.

If $a+1 = 5$, then $b+1 = 2$ or $4$. The first case gives $n = 2^4 \cdot 5^1 \cdot p$ where $p$ is a prime other than 2 or 5. Thus we have $80p < 2019 \implies p = 3, 7, 11, 13, 17, 19, 23$. The sum of all such $n$ is $80(3+7+11+13+17+19+23) = 7440$. In the second case $b+1 = 4$ and $d(k) = 1$, and there is one solution $n = 2^4 \cdot 5^3 = 2000$.

If $a+1 = 10$, then $b+1 = 2$, but this gives $2^9 \cdot 5^1 > 2019$. No other values for $a+1$ work.

Then we have $\frac{S}{20} = \frac{80(3+7+11+13+17+19+23) + 2000}{20} = 372 + 100 = \boxed{472}$.

-scrabbler94

Solution 2

For $n$ to be 20-pretty, $n$ can only take on certain prime factorization forms: namely, $p^{19}, p^9q, p^4q^3, p^4qr$. Notice that for the first form, the smallest possible integer with it is $2^{19}$ and the smallest integer with the second form divisible by 20 is $2^{9}5$ which are both greater than 2019.

For the third form, the only numbers with it and divisible by 20 is $2^45^3=2000$ and $2^35^4=5000$ so only $2000$ is 20-pretty with this factorization.

For the fourth form, $2^45r$ is the sufficient combination for $20|n$. Since $n=80r<2019$, $r\le 25$. Therefore, $r$ can take on $3, 7, 11, 13, 17, 19,$ or $23$.

Thus, $\frac{S}{20}=\frac{2000+80(3+7+11+...+23)}{20}=100+4(3+7+11+...+23)=\boxed{472}$.

See Also

2019 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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